Question: What is the slope of the line tangent to $f(x) = -x^{2}-2x+3$ at $x = -3$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-(x+h)^{2}-2(x+h)+3) - (-x^{2}-2x+3)}{h}$ $ = \lim_{h \to 0} \frac{(-(x^{2}+2x h+h^{2})-2(x+h)+3) - (-x^{2}-2x+3)}{h}$ $ = \lim_{h \to 0} \frac{-x^{2}-2(x h)-h^{2}-2x-2h+3+x^{2}+2x-3}{h}$ $ = \lim_{h \to 0} \frac{-2(x h)-h^{2}-2h}{h}$ $ = \lim_{h \to 0} -2x-h-2$ $ = -2x-2$ $ = (-2)(-3)-2$ $ = 4$